# If a,b are the roots ( assumed non 0 ) of 3x^2-4x+1=0 form the equation whose roots are a^2/b and b^2/a.

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3x^2 - 4x + 1 = 0

The roots are a and b

==> a+ b = 4/3.....(1)

==> a*b = 1/3......(2)

Now for the equation whose roots a^2/b and b^2/a

Sum = a^2/b + b^2/a = (a^2 + b^3)/ab

= (a+b)^3 - 3ab(a+b)/ab

= (a+b)^3 - 3(a+b)

From (1) and (2):

==> Sum = (4/3)^3 - 3 (4/3)

= 64/27 - 4 = 64- 108/ 4 = -44/4 = -11

==> Product= (a^2/b)*(b^2/a)= (ab)^2/ ab

= (ab) = 1/3

Then, the equation is:

x^2 + 11x + 1/3 = 0

Multiply by 3:

==> 3x^2 + 33x + 1 = 0

3x^2-4x+1 = 0

Let a and b be the roots of the equation.

Then 3x^2 -4x+1 = (x-a).

By the relation between roots and coefficients,

Sum of roots =a+b = -(-4/3) =4/3 and Product of roots.ab = 1/3.

Therefore sum of a^2/b and b^2 /a = (a^+b^3)/ab = {(a+b)^3-3(a+b)}/ab = ((4/3)^3-3(1/3)(4/3)}/(1/3) = 3{64/27 - 4/3} = 64/9-4 = (64-36)/9 = 28/9.

Product of a^2/b and b^2/a = (a^2*b^2)/ab = ab = 1/3.

Therefore the equation whose roots are a^2/b andb^2/a is

x^2 -(sum ofroots =28/9)x +(product of roots = 1/3) = 0

x^2-28x/9 +1/3 = 0. To getrid of fractional coefficients, multiply by 9.

9x^2-28x+3 = 0 is the required equation.

If a and b are the roots of the given equation, then, using Viete's relationships, we could write:

a+b = - (-4)/3

a*b = 1/3

Now, we know that we could write an equation if we know the sum and the product of it's roots:

x^2 - Sx + P = 0

We know that the roots of the equation we have to form are: a^2/b and b^2/a.

We'll write the sum of them:

S = a^2/b + b^2/a = (a^3 + b^3)/a*b

But the sum of cubes from numerator could be written as:

a^3 + b^3 = (a+b)^3 - 3a*b(a+b)

We'll substitute in the relation above, the values of a+b and a*b:

(a+b)^3 - 3a*b(a+b) = (4/3)^3- 3*(1/3)(4/3) = (4/3)(16/9 - 1)

a^3 + b^3 = (4/3)*(7/9) = 28/27

S = (28/27)/(1/3)

**S = 28/9**

**P = (a^2/b)* (b^2/a) =a*b = 1/3**

The quadratic equation whose roots are (a^2/b) and (b^2/a) is:

x^2 - (28/9)*x + 1/3 = 0

**9x^2 - 28x + 3 = 0**